Compute all possible lengths of sides AB and AC

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Compute all possible lengths of sides AB and AC

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#1 h Oct 31, 2010, 1:08 PM • 1 Y

Y by Adventure10

In the triangle $ABC$ , the angle $\alpha = \angle BAC$ and the side $a = BC$ are given. Assume that $a = \sqrt{rR}$ , where $r$ is the inradius and $R$ the circumradius. Compute all possible lengths of sides $AB$ and $AC.$

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#2 h Jul 31, 2014, 7:51 PM • 7 Y

Y by Adventure10, Mango247, and 5 other users

Let $\Delta$ denote the area of $\triangle{ABC}$ . Using $\Delta=\frac{abc}{4R}=rs$ we get $b+c=\frac{bc-2rR}{2\sqrt{rR}}\quad (\diamond)$ after some simple manipulation. Now by the Cosine rule, $b^2+c^2-2bc\cos \alpha = a^2$ or $(b+c)^2-2bc(1+\cos \alpha)=rR$ . Plugging in our equation for $b+c$ and letting $t:=bc$ we get $\left(\frac{t-2rR}{2\sqrt{rR}}\right)^2-2t(1+\cos \alpha)-rR=0$ or $\frac{t^2}{4rR}+rR-t-2t(1+\cos \alpha)-rR=0$ or $t^2-4rR(3+2\cos \alpha)t=0$ . Since $t=bc\ne 0$ we get divide by it obtaining $t=4rR(3+2\cos\alpha)$ . Finally, feeding this and $c=\frac{4rR}{b}(3+2\cos\alpha)$ into $(\diamond)$ we can get a quadratic in $b$ , then we solve for $b$ and make a back-substitution to get $c$ . (The last part is really bashy, best avoid it.)

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